Question

If only 0.225 g of Ca(OH)2 dissolves in enough water to give 0.100 L of aqueous solution at a given temperature, what is the Ksp value for calcium hydroxide at this temperature

Answer 1

Answer: The $K_{sp}$ value for calcium hydroxide at this temperature is $1.08 \times 10^{-4}$.

Explanation:

Given: Mass of $Ca(OH)_{2}$ = 0.225 g

Volume = 0.100 L

As moles is the mass of substance divided by its molar mass.

So, moles of $Ca(OH)_{2}$ (molar mass = 74 g/mol) is calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{0.225 g}{74 g/mol}\\= 0.003 mol$

Molarity is the number of moles of substance present in a liter of solution.

Hence, molarity of given solution will be as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.003 mol}{0.1 L}\\= 0.03 M$

The equation for dissociation of  $Ca(OH)_{2}$ is as follows.

$Ca(OH)_{2} \rightarrow Ca^{2+} + 2OH^{-}$

This means that $[Ca^{2+}] = 0.03$ and $[OH^{-}] = 2 \times 0.03 = 0.06$. Hence, $K_{sp}$ value for this reaction is calculated as follows.

$K_{sp} = [Ca^{2+}][OH^{-}]^{2}\\= (0.03) \times (0.06)^{2}\\= 1.08 \times 10^{-4}$

Thus, we can conclude that the $K_{sp}$ value for calcium hydroxide at this temperature is $1.08 \times 10^{-4}$.